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Weird Mathematics And The Planets
By
Ian Beardsley
Copyright © 2020 by Ian Beardsley!
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!
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It is as if the planets interior to the asteroid belt are distributed by doing what I call weird
calculus. And, that the planets exterior to the asteroid belt are doing normal calculus. It is as if
the planets interior to the asteroid belt are trying to take the derivative of x to the n without
using logarithms. There seems to be a connection between artificial intelligence elements and
the planets.!
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Table of Contents
Quantization of Planetary Orbits In Terms of AI Binary……….5"
The Conundrum……………………………………………………7"
The Weird Calculus………………………………………………….8
Understanding The Weird Calculus………………………………….9
The Golden Ratio…………………………………………………….12
Weird Arithmetic……………………………………………………..15
Mathematical Look At Weird Calculus……………………………….24
Silicon and Carbon……………………………………………………32
Germanium and Carbon………………………………………………33
The Scheme…………………………………………………………..35
The Planets and AI……………………………………………………40
The Protoplanetary Disc………………………………………………52
Bone…………………………………………………………………..55
Boron………………………………………………………………….67
Masculine And Feminine……………………………………………..85
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Quantization of Planetary Orbits In Terms of AI Binary
I have devised a scheme for the planets in terms of the golden ratio conjugate phi and Euler’s
number e:"
"
"
"
"
"
"
"
So that for the planets exterior to the asteroid belt or where
is is the asteroid belt ( )
Which is the solution to the differential equation
Where have we seen this? In computer science.
means
Where n is the number of bits in a number N in binary. We write in binary
0=0
1=1
10=2
11=3
100=4
101=5
110=6
(1 ϕ)e
ϕ
= 0.7AU = Venus
ϕe
(1ϕ)
= 0.9AU = Ear th
ϕ
2
e
(2ϕ)
= 1.52 = M ars
2ϕe
(2ϕ)
= 4.9 = Jupiter
4ϕe
(2ϕ)
= 10 = Sat ur n
16ϕe
(2ϕ)
= 39.38 = Neptune
P
n
= 2
n
ϕe
(2ϕ)
P
n
= c2
n
c = ϕe
(2ϕ)
= 2.461
C 2
0
= C = 2.461
P
0
P
1
= Jupiter
P
2
= Sat ur n
P
3
= Ura nu s
P
4
= Nept u ne
d
2
y
dn
2
2log(2)
dy
dn
+ log
2
(2)y = 0
log
2
N = n
2
n
= N
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111=7
1000=8
1001=9
1010=10
1011=11
1100=12
1101=13
1110=14
1111=15
10000=16…
But what is interesting about this?
You can’t have a fractional number of bits, thus the spectrum is quantized according to whole
number solutions of
But so are the planets given by
"
"
"
"
Meaning, since we have 2, 4, 8. 16 that the planets are quantized into whole number orbits
according to computer binary with Jupiter as 2, Saturn as 4, Uranus as 8, and Neptune as 16 if
we do it in terms of Euler’s number, e and the golden ratio conjugate, ."
That is, 2=10, 4=100, 8=1000, 16=10000"
Are all zeros after a one.!
log
2
3 = n
n =
log3
log2
= 1.5847
2
n
= N
P
n
= c2
n
2ϕe
(2ϕ)
= 4.9 = Jupiter
4ϕe
(2ϕ)
= 10 = Sat ur n
16ϕe
(2ϕ)
= 39.38 = Neptune
ϕ
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The Conundrum
It is as if the planets interior to the asteroid belt are distributed by doing what I call weird
calculus. And, that the planets exterior to the asteroid belt are doing normal calculus. It is as if
the planets interior to the asteroid belt are trying to take the derivative of x to the n without
using logarithms. This in the sense that:"
If we refer back to the foundations of calculus, while the integral of simple functions can be
considered
We have a conundrum for
That the power rule gives:
Thus to get around this, we searched for a function such that the integral holds, and as such we
discovered the natural logarithm (ln) and Eulers number e. And we have
Where
And, the derivative of is itself and e is the transcendental and irrational number given by
e=2.718…
That is, while
x
n
d x =
x
n+1
n + 1
+ C
f (x) =
1
x
= x
1
1
x
d x =
x
1+1
0
1
x
d x = ln(x) + C
d
dx
e
x
= e
x
ln(x) = log
e
(x)
e
x
f
1
ln(x)
1
ln(x)
f
1
ln(x) = e
x
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We can approximate any function with a polynomial, the simplest example being the linear
approximation formed by writing the change in f(x) due to a change in x:
This results in Taylors formula
From which we derive the Taylor series
We know the kth derivative of e to the x is e to the x itself. Thus,
The Weird Calculus
The planets seem to the think
But, rather is
f (x) = f (a) + f (a)(x a)
f (x) = f (a) + f (a)(x a) +
f (a)
2!
(x a)
2
+
n=0
f
n
(a)
n!
(x a)
n
= f (a) + f (a)(x a) +
f (a)
2!
(x a)
2
+
f
(k)
(x) = e
x
e
x
= 1 + x +
x
2
2!
+
x
3
3!
+
lim
n→∞
x
n
n!
= 0
e
x
=
n=0
x
n
n!
= 1 + x +
x
2
2!
+
x
3
3!
+
e =
n=0
1
n!
= 1 +
1
1!
+
1
2!
+
1
3!
+ = 2.718
d
dx
e
x
e
x
d
dx
e
x
= xe
(x1)
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makes me think of the derivative of x to the n:"
"
And so does"
"
Because we have"
yields"
by way of"
= "
And, the is like ."
And,…"
makes me think of the second derivative of x to the n:"
"
That is:"
"
"
Because we have"
"
Thus, in general the derivative in weird calculus is:"
"
Understanding The Weird Calculus
The plot of has got to be one of the most interesting things I have ever seen:"
(1 ϕ)e
ϕ
d
dx
x
n
= n x
(n1)
ϕe
(1ϕ)
(n 1)
(1 ϕ)
(ϕ 1)
(1 ϕ)
ϕe
n x
ϕ
2
e
(2ϕ)
d
2
dx
2
x
n
= n(n 1)x
(n2)
d
d x
n x
(n1)
= n(n 1)x
(n11)
= n(n 1)x
(n2)
(n 2)
(ϕ 2) = (2 ϕ)
k
th
d
(k)
y
dn
(k)
= n
k
x
(nk)
n x
n
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!
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Thus in regular calculus"
"
And, in weird calculus"
"
This gives"
"
"
Let us compare regular calculus to weird calculus:"
"
If we take"
="
1.10428"
Where"
and has the property ."
But if we use weird calculus to take the second derivative (written respect to x for weird
derivative)"
"
It is exactly the Mars orbit."
And, if we simply take, we get…"
=8.15956"
d
dx
e
x
= e
x
d
dx
e
x
= xe
(x1)
d
dϕ
e
ϕ
= ϕe
(ϕ1)
= 0.9055
d
dϕ
e
ϕ
= e
ϕ
= 1.855
1.855
0.9055
= 2.04859 2
d
dϕ
ϕe
(ϕ1)
= e
(ϕ1)
(ϕ + 1) = e
(ϕ1)
1.618
Φ =
1
ϕ
Φ = ϕ + 1
d
2
dx
2
e
ϕ
= ϕ
2
e
(ϕ2)
= 1.52
d
dϕ
ϕe
ϕ+1
= e
ϕ+1
(ϕ + 1) = e
Φ
(ϕ + 1) = e
Φ
Φ
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I think the planets do weird calculus because it is a doubling eect in that"
"
Because it keeps the planets from interfering with one another so they don’t get torn apart as
they did with the asteroid belt. The second derivative of e to the phi is itself so the 1.855 is
constant. Comparing this to the second derivative of weird calculus we have:"
"
Notice that"
"
The Golden Ratio
But what is the golden ratio and its conjugate ."
We can derive the golden ratio as such (refer to fig 13):"
"
"
"
"
"
"
"
"
1.855
0.9055
= 2.04859 2
1.855
1.52
= 1.220
1.22
2
= 0.61 ϕ
Φ
ϕ
a
b
=
b
c
= Φ
a = b + c
ac = b
2
c = a b
a(a b) = b
2
a
2
ab b
2
= 0
a
2
b
2
a
b
1 = 0
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"
"
"
"
"
"
"
Let us say a/b=x, the golden ratio. Then,…"
"
Let us dierentiate this implicitly:"
"
"
"
Which is similar to Euler’s number, e because it is the base such that is itself :"
"
But "
a
2
b
2
a
b
+
1
4
= 1 +
1
4
(
a
b
)
2
a
b
+
1
4
= 1 +
1
4
(
a
b
)
2
a
b
+
1
4
=
4
4
+
1
4
=
5
4
(
a
b
1
2
)
2
=
5
4
a
b
1
2
=
5
2
Φ =
5 + 1
2
ϕ =
b
a
=
5 1
2
x
2
x 1 = 0
d
dx
x
2
d
dx
x
d
dx
1 = 0
2x 1 = 0
x =
1
2
d
dx
e
x
e
x
d
dx
e
x
= e
x
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"
Which says for this angle the x-component equals the y-component is that is , x=1/2
bisects a right angle. Which similar in concept to Euler’s number e because it is the base such
that is itself . But if , then:"
"
It is the diagonal of the unit square. We notice something interesting happens:"
"
, , "
Where is the cosine of 30 degrees, in the unit equilateral triangle in which the altitude has
been drawn in (fig 14):"
!
sin 45
= cos45
=
2
2
1
2
90
d
dx
e
x
e
x
sin 45
= cos45
=
2
2
2 cos
π
4
= 2
2 cos
π
n
=
2 cos
π
4
= 2
2 cos
π
5
= Φ
2 cos
π
6
= 3
3
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Weird Arithmetic
Characterizing the distribution of the planets around the sun seems to defy a mathematical
expression. Even the Titius-Bode rule falls apart pretty badly at Neptune."
The Titius-Bode Rule is:"
"
"
Which produces the orbits of the planets in astronomical units as such in AU:"
!
r = 0.4 + (0.3)2
n
n = ,0,1,2,…
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However, I find if we break-up the solar system into two parts; planets interior to the asteroid
belt, and planets exterior to the asteroid belt, quite an interesting pattern forms:!
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Thus we can define a weird arithmetic too. We begin by changing the order of operations and
say that:"
"
So that"
2n-3=2(2-3)=2(-1)=-2"
We see that since in normal math"
(1)x=x"
"
Is the multiplicative identity and"
x+0=x"
Is the additive identity that in weird math"
-1/4=4, -1/2=2,…"
Thus,…"
2(1-3)=-4=1/4=0.25=mercury=0.4 AU"
2(2-3)=-2=1/2=0.5=venus=0.72 AU"
2(3-3)=0.0=1=Earth=1.00 AU"
(In weird math zero the additive identity is the 1 the multiplicative identity. Which, resolves the
enigma of infinity: which says because infinity is
now 1/0=1/1=1 infinity is 1, a whole encompassing any multiplicative of it.)"
2(4-3)=2=Mars=1.52 AU"
2(5-3)=4=Asteroids=2-2AU"
2(6-3)=6=Jupiter=5.2 AU"
2(7-3)=8=Saturn=9.5 AU"
2(8-3)=10=Uranus=19 AU"
2(9-3)=12=Neptune=30 AU"
Then we establish the connection between regular math and weird math by taking "
"
Which has plots:!
2n 3 = 2(n 3)
1/0 = 2/0,3/0,… =
= 2 = 3
2n 3
2n 6
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!
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Its Taylor Expansion is:"
"
Which approaches the orbits of the first five planets including the asteroid belt if we take each
term separately:"
"
"
"
"
"
"
Then after the asteroids it skips to n to the sixth for Jupiter:"
"
"
"
"
Thus, the after the asteroid the exponent of n counts 6, 9, 11,12 which is"
9-6=3 and 11-9=2 and 12-11=1 producing"
3, 2, 1"
2n 3
2n 6
=
1
2
n
6
n
2
18
n
3
54
n
4
162
+ O(n
5
)
n = 18 = 4.24264
1
2
= 0.5 = mercur y = 0.4AU
n
6
= 0.7 = venus = 0.72AU
n
2
18
= 1.00 = ear th = 1.00AU
n
3
54
= 1.412 2 = m ars = 1.52AU
n
4
162
= 1.999 2 = asteroids = 2AU 3AU
n
6
1458
= 5.1 = jupiter = 5.2AU
n
9
39366
= 11.31AU = sat ur n = 9.5AU
n
11
354294
= 22.624AU = ura nus = 19AU
n
12
1062882
= 32AU = neptune = 30AU
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Thus the equation for the planets is:"
"
Which is"
"
i=(1, 2, 3, 4, 5..)"
"
"
."
."
."
If we write it"
"
i=(0, 1, 2, 3, 4…)"
x=(1, 2, 3, 4, 5..)"
"
"
."
."
."
The plots are…!
(
1
2
,
n
6
,
n
2
18
,
n
3
54
,
n
4
162
)
P
i
=
(
1
2
,
n
i
2 3
i
, . . .
)
P
1
=
n
1
2 3
1
=
n
6
P
2
=
n
2
2 3
2
=
n
2
18
P
i
=
(
1
2
,
n
i+1
2 3
x
)
P
0
=
n
0+1
2 3
1
=
n
6
P
1
=
n
1+1
2 3
2
=
n
2
18
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!
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We can produce the exponents for the planets with a short piece of code. The idea that we can
model reality with a few short lines of code instead of algebra, comes from Wolfram. If you
know C++, here is the code"
//
// main.cpp
// planets
//
// Created by Ian Beardsley on 12/4/20.
// Copyright © 2020 Ian Beardsley. All rights reserved.
//
#include <iostream>
int main(int argc, const char * argv[])
{
int i=0;
{
printf("0\n");
while (i!=4)
{
i=i+1;
printf("%i\n",i);
}
int i=4;
printf("\n");
printf("\n");
while (i!=1)
{
i=i-1;
printf("%i\n",i);
}
}
return 0;
}
And Running the Code we have:"
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/Users/beardsleyian/Desktop/planets ; exit;
Ians-MBP:~ beardsleyian$ /Users/beardsleyian/Desktop/planets ; exit;
0
1
2
3
4
3
2
1
logout
Saving session...
...copying shared history...
...saving history...truncating history files...
...completed.
of 24 88
Mathematical Look At Weird Calculus!
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Referring to my paper Weird Mathematics and The Planets (Beardsley, 2020), the weird
derivative was provoked by the planets, but I feel one should look at it mathematically. While I
will do this time permitting I would just like to present the immediate relationships between
y=f(x) and its weird derivative for f(x) equal to e to the x, and not put it in the paper until I can
go into it more extensively."
We look at the plot of e to the x whose derivative is e to the x:!
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Then we look at the plot of the weird derivative of e to the x and get:"
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We see its maximum is not to pass one and that its limit as x goes to infinity is zero, and that
the graph of the e to the x and its weird derivative on the same graph speak of the graph of the
tangent function:"
"
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Thus we put the tangent function in the graph of both e to the x and its weird derivative and we
see it more clearly:"
"
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This is interesting because tangent being y/x is similar to the slope of the tangent, the
derivative (dy/dx) hence the terminology that the the derivative of y=f(x) evaluated at x is the
slope of the tangent of f(x) at (x, y). To see the meaning of this we plot by hand and make the
necessary calculations:"
!
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"
Where the plot labeled tan(x) is really tan(x)+1/2 which is what I thing we should really be
looking at. Which is pivotal if we consider what we wrote earlier:"
Let us say a/b=x, the golden ratio. Then,…"
"
Let us dierentiate this implicitly:"
"
x
2
x 1 = 0
d
dx
x
2
d
dx
x
d
dx
1 = 0
of 31 88
"
"
Which is similar to Euler’s number, e because it is the base such that is itself :"
"
But "
"
Which says for this angle the x-component equals the y-component is that is , x=1/2
bisects a right angle. Which similar in concept to Euler’s number e because it is the base such
that is itself . But if , then:"
"
It is the diagonal of the unit square. We notice something interesting happens:"
"
, , "
Where is the cosine of 30 degrees, in the unit equilateral triangle in which the altitude has
been drawn in (fig 14):"
!
2x 1 = 0
x =
1
2
d
dx
e
x
e
x
d
dx
e
x
= e
x
sin 45
= cos45
=
2
2
1
2
90
d
dx
e
x
e
x
sin 45
= cos45
=
2
2
2 cos
π
4
= 2
2 cos
π
n
=
2 cos
π
4
= 2
2 cos
π
5
= Φ
2 cos
π
6
= 3
3
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Silicon and Carbon
The golden ratio and the golden ratio conjugate are the solution of the quadratic"
that meets the conditions and a=b+c"
Where and , ."
We guess that artificial intelligence (AI) has the golden ratio, or its conjugate in its means
geometric, harmonic, and arithmetic by molar mass by taking these means between doping
agents phosphorus (P) and boron (B) divided by semiconductor material silicon (Si) :"
"
"
"
Which can be written"
"
We see that the biological elements, H, N, C, O compared to the AI elements P, B, Si is the
golden ratio conjugate (phi) as well:"
"
So we can now establish the connection between artificial intelligence and biological life:"
"
Which can be written:"
(
a
b
)
2
a
b
1 = 0
a
b
=
b
c
Φ =
a
b
ϕ =
5 1
2
ϕ =
1
Φ
PB
Si
=
(30.97)(10.81)
28.09
= 0.65
2PB
P + B
1
Si
=
2(30.97)(10.81)
30.97 + 10.81
1
28.09
= 0.57
0.65 + 0.57
2
= 0.61 ϕ
PB(P + B) + 2PB
2(P + B)Si
ϕ
C + N + O + H
P + B + Si
ϕ
(P + B + Si )
PB(P + B) + 2PB
2(P + B)Si
(C + N + O + H )
of 33 88
"
Where HNCO is isocyanic acid, the most basic organic compound. We write in the arithmetic
mean:"
"
Which is nice because we can write in the second first generation semiconductor as well
(germanium) and the doping agents gallium (Ga) and arsenic (As):"
"
Where"
"
Where ZnSe is zinc selenide, an intrinsic semiconductor used in AI, meaning it doesn’t require
doping agents. We now have:"
"
Germanium And Carbon
We could begin with semiconductor germanium (Ge) and doping agents gallium (Ga) and
Phosphorus (P) and we get a similar equation:"
, "
In grams per mole. Then we compare these molar masses to the molar masses of the
semiconductor material Ge:"
"
"
Then, take the arithmetic mean between these:"
PB
[
P
Si
+
B
Si
+ 1
]
+
2PB
P + B
[
P
Si
+
B
Si
+ 1
]
2HCNO
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
3HNCO
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
Zn
Se
[
P
Si
+
B
Si
+ 1
]
[
Ga
Ge
+
As
Ge
+ 1
]
PB
(
Zn
Se
)
+
2PB
P + B
(
Zn
Se
)
+
P + B
2
(
Zn
Se
)
HNCO
2Ga P
Ga + P
= 42.866
Ga P = 46.46749
2Ga P
Ga + P
1
Ge
=
42.866
72.61
= 0.59
Ga P
1
Ge
=
46.46749
72.61
= 0.64
of 34 88
"
We then notice this is about the golden ratio conjugate, , which is the inverse of the golden
ratio, . . Thus, we have"
1. "
2. "
This is considering the elements of artificial intelligence (AI) Ga, P, Ge, Si. Since we want to find
the connection of artificial intelligence to biological life, we compare these to the biological
elements most abundant by mass carbon (C), hydrogen (H), nitrogen (N), oxygen (O),
phosphorus (P), sulfur (S). We write these CHNOPS (C+H+N+O+P+S) and find:"
"
A similar thing can be done with germanium, Ge, and gallium, Ga, and arsenic, As, this time
using CHNOPS the most abundant biological elements by mass:"
"
"
"
"
"
We can also make a construct for silicon doped with gallium and phosphorus:"
"
0.59 + 0.64
2
= 0.615
ϕ
Φ
ϕ
1
Φ
Ga P(G a + P) + 2G aP
2(Ga + P)Ge
ϕ
Ga P(G a + P) + 2G aP
2(Ga + P)Si
Φ
CHNOPS
Ga + As + Ge
1
2
[
Ga As +
2Ga As
Ga + As
+
Ga + As
2
][
Ga
Ge
+
As
Ge
+ 1
]
CHNOPS
[
Ga
Si
+
As
Si
+ 1
]
Ga As
(
O
S
)
+
2Ga As
Ga + As
(
O
S
)
+
Ga + As
2
(
O
S
)
CHNOPS
O
S
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga
Si
+
As
Si
+ 1
]
Ga As(G a + As) + 2Ga As
2(Ga + As)Ge
1
C + H + N + O + P + S
Ga + As + Ge
1
2
(C + N + O + H )
2(Ga + P)Si
Ga P(G a + P) + 2G aP
(P + B + Si )
of 35 88
"
"
And for germanium doped with gallium and phosphorus:"
"
"
"
Now we develop a scheme for naming the AI elements with numbers that might reveal
patterns."
The Scheme
Silicon and germanium are in group 14 like carbon (C) and as such have 4 valence electrons. Thus to have
positive type silicon and germanium, they need doping agents from group 13 (three valence electrons)
like boron and gallium, and to have negative type silicon and germanium they need doping agents from
group 15 like phosphorus and arsenic.
We need a scheme that takes all of this into account.We begin with the periodic table of the
elements:"
"
HNCO
2(Ga + P)Si
(Ga + P)
[
Ga P +
2GaP
Ga + P
]
(P + B + Si )
HNCO
2(P + B + Si )Si
Ga P +
2GaP
Ga + P
Ga P(G a + P) + 2G aP
2(Ga + P)Ge
ϕ
[
Ga P +
2Ga P
Ga + P
+
Ga + P
2
][
P
Ge
+
B
Ge
+
Si
Ge
]
HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
Ga P
(
B
S
)
+
2Ga P
Ga + P
(
B
S
)
+
Ga + P
2
(
B
S
)
HNCO
of 36 88
And then pull out that section with the primary artificial intelligence (AI) elements:"
And we next number the elements as such,…"
Thus 13 is an element in row 1 and must be boron because it has three valence electrons. 14 is carbon
because it is in row 1 and has four valence electrons. And so on. So, instead of element Si we have E_24,
and instead of element Ge, we have E_34, and so on. Thus,…
Becomes,…"
"
"
of 37 88
"
Where E means “element”. Now we define operators for the geometric, harmonic, and
arithmetic means:"
"
"
"
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
̂
G = a b
̂
H =
2ab
a + b
̂
A =
a + b
2
of 38 88
And our equation: Becomes:"
"
And Becomes"
"
The full set of AIbioequations using the standard notation is:!
PB(P + B) + 2PB
2(P + B)Si
ϕ
̂
G(E
25
, E
13
) +
̂
H(E
25
, E
13
)
2E
24
ϕ
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
[
̂
G(E
25
, E
13
) +
̂
H(E
25
, E
13
) +
̂
A(E
25
, E
13
)
]
[
E
25
E
24
+
E
13
E
24
+ 1
]
HNCO
[
E
33
E
34
+
E
35
E
34
+ 1
]
of 39 88
"
"
"
"
"
"
"
"
"
"
"
!
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga As +
2Ga As
Ga + As
+
Ga + As
2
][
Ga
Ge
+
As
Ge
+ 1
]
CHNOPS
[
Ga
Si
+
As
Si
+ 1
]
[
Ga P +
2Ga P
Ga + P
+
Ga + P
2
][
P
Ge
+
B
Ge
+
Si
Ge
]
HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
HNCO
2(P + B + Si )Si
Ga P +
2GaP
Ga + P
PB(P + B) + 2PB
2(P + B)Si
ϕ
Ga As(G a + As) + 2Ga As
2(Ga + As)Ge
1
Ga P(G a + P) + 2G aP
2(Ga + P)Ge
ϕ
Ga P(G a + P) + 2G aP
2(Ga + P)Si
Φ
C + N + O + H
P + B + Si
ϕ
C + H + N + O + P + S
Ga + As + Ge
1
2
Zn
Se
[
P
Si
+
B
Si
+ 1
]
[
Ga
Ge
+
As
Ge
+ 1
]
O
S
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga
Si
+
As
Si
+ 1
]
of 40 88
The Planets and AI!
of 41 88
However, if we leave the microscopic world and go to the macroscopic, the planets, we have
the the 3/2 appears in the ratio of the terrestrial planets Earth, and Mars, which is 1.52 which I
said was in earlier work was:"
"
However, we used values of Si=28.09 and Ge=72.61. Recent measurements of Ge give a
slightly modified value for Ge. It is 72.64. Let us compute these two equations since they are,
as we are guessing the most important, in terms of Si, to two places after the decimal, and for
Ge using the most recent value:"
Ge=72.64 g/mol"
Si=28.0855 g/mol"
Thus,…"
"
We see it is still accurate to two places after the decimal. Let us now look at Mars,…"
"
It needs to be close to 1.52AU and we see we made a mistake with the equation to begin with.
We find the answer is in averaging this with the dierence of squares in the denominator with
the square of the dierence in denominator, that is"
with "
Which gives our modified Mars equation:"
"
We compute its accuracy:"
"
This has an accuracy of:"
, 97%"
1.52 =
2SiGe
Ge
2
Si
2
1
72.64
2
2(28.0855)(72.64) +
28.0855
3
72.64
1 +
28.0855
2
72.64
2
= 0.722995806
2SiGe
Ge
2
Si
2
=
2(28.0855)(72.64)
72.64
2
28.0855
2
= 0.909
2SiGe
Ge
2
Si
2
2SiGe
(Ge Si)
2
m ars =
2SiGe
2
(Si Ge)
2
(Si + Ge)
m ars =
2(28.0855)(72.64)
2
199950.5396
= 1.48AU
1.48/1.52 = 0.973684
of 42 88
The plot is still like a propeller, but more stretched out. It was"
But, the modified Mars equation give us:"
Our updated AI planetary table becomes…!
of 43 88
"
of 44 88
We proceed to look at what the equation for the earth might look like (1.00 astronomical units).
It is is a related polynomial that followed in my mind from the above. I simply wrote the
denominator with the square of the dierences instead of the the dierence of the squares, and
did not multiply the product between Si and Ge by 2. It works nearly exact as well:"
"
Which can be written"
and has plot"
We look at this and guess Venus is:"
"
Or,"
"
It turns out it is the arithmetic mean between the two which is:"
"
1.01 =
SiGe
(Ge Si)
2
SiGe
(Ge
2
2SiGe + Si
2
)
2SiGe
Ge
2
+ Si
2
2SiGe
Ge
2
=
2Si
Ge
0.72 =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
of 45 88
And has plot"
All of this using Si=28.09 g/mol, and Ge=72.61g/mol!
of 46 88
Mars "
"
Earth "
"
1.48 =
2SiGe
2
(Si Ge)
2
(Si + Ge)
1.01 =
SiGe
(Ge Si)
2
of 47 88
Venus !
0.72 =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
of 48 88
Earth equation!
of 49 88
I said the Venus orbit was "
"
And, that I arrived at it by averaging my two estimates, which were"
and,…"
"
It is then simple to attain the equation for mercury because its average distance from the sun is
0.4 astronomical units, but since it has a highly eccentric orbit it can drop to below this or go
above it. The orbit can be described then by one half either one of these equations. Their plots
are:"
!
0.72 =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
2SiGe
Ge
2
+ Si
2
2SiGe
Ge
2
=
2Si
Ge
of 50 88
"
"
Now we move to the outer planets, most of which are gas giants. It becomes obvious what we
do for Jupiter, which is at 5.2 AU from the sun on the average. We take the earth equation:"
"
And, invert it, put the polynomial in the numerator, and the product in the denominator, but to
follow previous patterns, reverse the negative sign and we have:"
"
It is incredible how close it is to 5.2 AU, and it can be made to work perfectly by adjusting the
coecients of the polynomial, however we want as above because we want the polynomial in
the numerator to be a perfect square. That is, it can be factored"
"
Its plot is"
"
SiGe
Ge
2
+ Si
2
= 0.3
SiGe
Ge
2
=
Si
Ge
= 0.387
SiGe
(Ge
2
2SiGe + Si
2
)
= 1.01
Si
2
+ 2SiGe + Ge
2
SiGe
= 4.97AU 5AU
(Si + Ge)
2
SiGe
= 4.97
of 51 88
We now move out to Saturn and the solution is simple. Just as the Mercury equations were half
the Venus equations, Saturn (9 AU) is twice the Jupiter equation:"
"
Which of course, like the Jupiter equation can have its coecients adjusted. It plot is:"
"
The next planet out, Uranus (19.2 AU), is clearly about twice the Saturn Equation:"
"
And, Neptune (30.33 AU) is clearly three times the Saturn Equation:"
"
Pluto has been declassified as a planet."
2(Si + Ge)
2
SiGe
= 9.94
4(Si + Ge)
2
SiGe
= 19.88
6(Si + Ge)
2
SiGe
= 29.82
of 52 88
The Protoplanetary Disc
If our protoplanetary disc, as I suggested, can be taken as having a molar mass that is the
mean between germanium and silicon, then it can be taken as having a density that is the
mean between the density of germanium and silicon."
!
of 53 88
"
Taking the protoplanetary disc as a thin disc we integrate from its center to the edge, with
density decreasing linearly to zero at the edge. Thus, if the density function is given by"
"
And, our integral is"
"
"
"
The mass of the solar system adding up all the planets yields"
"
That accounts for"
82% of the mass of the solar system not including the sun, that is, of the
protoplanetary disc surrounding the sun."
Using germanium alone, we get,"
"
If we weight the mixture of silicon and germanium as 1/3 and 2/3, then we have"
"
Which is very close."
93%"
This is all very good, because I only used the planets and asteroids."
Si + Ge
2
=
2.33 + 5.323
2
= 3.8265g/cm
3
ρ(r) = ρ
0
(
1
r
R
)
M =
2π
0
R
0
ρ
0
(
1
r
R
)
rdrdθ
M =
πρ
0
R
2
3
π(3.8265)(7.4 × 10
14
)
2
3
= 2.194 × 10
30
grams
M = 2.668 × 10
30
grams
2.194
2.668
100 =
π(5.323)(7.4 × 10
14
)
2
3
= 3.05 × 10
30
grams
π(4.32467)(7.4 × 10
14
)
2
3
= 2.48 × 10
30
grams
2.48
2.668
100 =
of 54 88
Weighting silicon and germanium as 1/4 and 3/4 we have"
"
Which accounts for"
98%"
Of the mass of the solar system (very accurate)."
This mixture of 1/4 to 3/4 is a combination that exists in the Earth atmosphere which is
approximately the mixture of oxygen to nitrogen. The earth atmosphere can be considered a
mixture of chiefly O2 and N2 in these proportions:"
Air is about 25% oxygen gas (O2) by volume and 75% nitrogen gas (N2) by volume meaning
the molar mass of air as a mixture is:"
"
By molar mass the ratio of air to H20 (water) is about the golden ratio:"
"
I am not saying the solar system is a thin disk with density of the weighted mean somewhere
between silicon and germanium, but that it can be modeled as such, though if the
protoplanetary disk that eclipses epsilon aurigae every 27 years is any indication of what a
protoplanetary cloud is like, it is a thin disk in the sense that it is about 1 AU thick and 10 AU in
diameter. This around a star orbiting another star."
π(4.4.57475)(7.4 × 10
14
)
2
3
= 2.623 × 10
30
grams
2.623
2.668
100 =
0.25O
2
+ 0.75N
2
air
air
H
2
O
Φ
of 55 88
Bone!
of 56 88
Part 1: Earlier Work"
It would seem there is some possibility that as life goes from the fundamental framework to
form, so does the math that describes it. I make no attempt to understand why the
composition of life is aesthetically pleasing and how that came about, but I think it is due to the
need for function. The structure herein found arose when I was comparing biological life to
artificial intelligence. The case for bone was so interesting to me that I decided to proceed to
muscle and skin. It is because of this structure only revealing itself when comparing the
biological to AI, that I might suggest you can only speak about what life is relative to another
construct, like AI. Especially where awareness is concerned if we consider the Turing test.!
of 57 88
In my exploration of the connection between biological life and AI the most dynamic
component is that of bone. It aords us the opportunity to look at:"
Multiplying Binomials"
Completing The Square"
The Quadratic Formula"
Ratios"
Proportions"
The Golden Ratio"
The Square Root of Two"
The Harmonic Mean!
of 58 88
Density of silicon is Si=2.33 grams per cubic centimeter."
Density of germanium is Ge=5.323 grams per cubic centimeter."
Density of hydroxyapatite is HA=3.00 grams per cubic centimeter."
This is"
where "
Where HA is the mineral component of bone, Si is an AI semiconductor material and Ge is an
AI semiconductor material. This means"
"
The harmonic mean between Si and Ge is HA,…"
"
This is the sextic,…"
"
Which has a solution"
"
Where x=Si, and y=Ge. It works for density and molar mass. It can be solved with the online
Wolfram Alpha computational engine. But,…"
"
"
!
3
4
Si +
1
4
Ge H A
HA = Ca
5
(PO
4
)
3
OH
Si
HA
Si +
[
1
Si
HA
]
Ge = HA
2SiGe
Si + Ge
HA
x
2
(x + y)
4
x y(x + y)
4
+ 2x y
2
(x + y)
3
4x
2
y
2
(x + y)
2
= 0
Si
Ge
=
1
2 + 1
1
HA
2
Si
2
Ge
HA
2
Si +
[
Ge
HA
1
]
= 0
Si =
1
2
Ge
±
HA
Ge
HA
2
4Ge
HA
+ 4
Si = Ge HA
of 59 88
"
"
"
"
"
Si
HA
Si +
[
1
Si
HA
]
Ge = HA
Si
2
HA
+ Ge
Si
HA
Ge H A
1
HA
Si
2
Ge
HA
Si + Ge HA
1
HA
2
Si
2
Ge
HA
2
Si +
Ge
HA
1
1
HA
2
Si
2
Ge
HA
2
Si +
Ge
HA
1 0
1
HA
2
Si
2
Ge
HA
2
Si +
[
Ge
HA
1
]
= 0
of 60 88
"
"
We see that the square of the binomial is a quadratic where the third term is the square of one
half the middle coecient. This gives us a method to solve quadratics called completing the
square:"
"
"
"
"
"
"
"
!
(x + a)(x + a) = x
2
+ 2a x + a
2
(x + a)
2
= x
2
+ 2a x + a
2
a x
2
+ bx + c = 0
a x
2
+ bx = c
x
2
+
b
a
x =
c
a
(
1
2
b
a
)
2
=
1
4
b
2
a
2
x
2
+
b
a
x +
1
4
b
2
a
2
=
c
a
+
1
4
b
2
a
2
(
x +
1
2
b
a
)
2
=
b
2
4ac
4a
2
x +
b
2a
=
±
b
2
4ac
2a
x =
b
±
b
2
4ac
2a
of 61 88
"
"
"
"
"
"
"
"
"
"
"
!
1
HA
2
Si
2
Ge
HA
2
Si +
[
Ge
HA
1
]
= 0
x =
b
±
b
2
4ac
2a
a =
a
HA
2
b =
Ge
HA
2
c =
[
Ge
HA
1
]
b
2
4ac =
Ge
2
HA
4
4
1
HA
2
[
Ge
HA
1
]
=
Ge
2
HA
4
4Ge
HA
3
+
4
HA
2
=
1
HA
2
[
Ge
2
HA
2
4Ge
HA
+ 4
]
b
2
4ac =
1
HA
(
Ge
HA
2
)
2
x =
Ge
HA
2
±
1
HA
[
Ge
HA
2
]
2
HA
2
=
1
2
Ge
±
1
2
HA
[
Ge
HA
2
]
=
1
2
Ge
±
1
2
Ge H A
Si =
1
2
Ge +
1
2
Ge H A
Si = Ge HA
of 62 88
"
"
"
"
"
"
"
"
"
"
"
"
"
"
"
Si Ge H A
HA
2SiGe
Si + Ge
Si Ge
2SiGe
Si + Ge
(Si + Ge)Ge
Si + Ge
(Si + Ge)Si
Si + Ge
2SiGe
Si + Ge
= 0
Ge
2
2SiGe Si
2
Si + Ge
= 0
x
2
2x y y
2
= 0
x
2
2x y = y
2
x
2
2x y + y
2
= 2y
2
(x y)
2
= 2y
2
x y =
±
2y
x = y + 2y
x = y(1 + 2)
x
y
= 1 + 2
y
x
=
1
2 + 1
Si
Ge
1
2 + 1
of 63 88
A ratio is and a proportion is which means a is to b as b is to c."
The Golden Ratio "
and. "
or "
"
"
"
"
"
"
"
"
a
b
a
b
=
b
c
(
Φ
)
a
b
=
b
c
a = b + c
ac = b
2
c =
b
2
a
a = b +
b
2
a
b
2
a
a + b = 0
b
2
a
2
1 +
b
a
= 0
(
b
a
)
2
+
b
a
1 = 0
(
b
a
)
2
+
b
a
+
1
4
= 1 +
1
4
(
b
a
+
1
2
)
2
=
5
4
b
a
=
1
2
±
5
2
b
a
=
5 1
2
a
b
=
5 + 1
2
ϕ =
5 1
2
Φ =
5 + 1
2
ϕ =
1
Φ
of 64 88
The mineral component of bone hydroxyapatite (HA) is"
"
The organic component of bone is collagen which is"
"
We have"
"
"
"
"
%"
"
!
Ca
5
(PO
4
)
3
OH = 502.32
g
mol
C
57
H
91
N
19
O
16
= 1298.67
g
mol
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
= 0.386795722
ϕ = 0.618033989
1 ϕ = 0.381966011
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
(1 ϕ)
0.381966011
0.386795722
100 = 98.75
Si
Ge
=
28.09
72.61
= 0.386861314 (1 ϕ)
Si
Ge
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
of 65 88
If Then, for any perfect square in general…
If… x=Si; B=Ge; A=HA we have…
We make the substitutions…
; ;
To get:
If ; ;
Then,…
If x=Si, B=Ge, A=HA then…
If k=4 then we have:
!
Si = Ge H A x = B A
1
HA
2
[
(
Ge
HA
)
2
k
Ge
HA
2
+
(
k
2
)
2
]
= b
2
4ac
1
A
2
[
(
B
A
)
2
k
B
A
2
+
(
k
2
)
2
]
= b
2
4ac
a =
k
4A
2
b =
B
A
2
c =
B
A
k
4
1
A
2
[
(
B
A
)
2
k
B
A
2
+
(
k
2
)
2
]
= b
2
4ac
a =
k
4A
2
b =
B
A
2
c =
B
A
k
4
k
4A
2
x
2
B
A
2
x +
[
B
A
k
4
]
= 0
k
4
1
HA
2
Si
2
Ge
HA
2
Si +
[
Ge
HA
k
4
]
= 0
1
HA
2
Si
2
Ge
HA
2
Si +
[
Ge
HA
1
]
= 0
of 66 88
If we want to turn Si=Ge-HA into
We write it:
Si+HA-Ge=0 and multiply through by Si:
Then, divide through by and get:
We see that
Thus we must multiply the third term in
By That is,…
Thus we define
and And we have…
Mapping our linear equation to a quadratic.
1
HA
2
Si
2
Ge
HA
2
Si +
[
Ge
HA
1
]
= 0
Si
2
+ HA Si Ge Si = 0
HA
2
1
HA
2
Si
2
Ge
HA
2
Si +
Si
HA
= 0
Si
HA
=
[
Ge
HA
1
]
1
HA
2
Si
2
Ge
HA
2
Si +
Si
HA
= 0
[
Ge
Si
HA
Si
]
[
Ge
Si
HA
Si
]
Si
HA
=
[
Ge
HA
1
]
̂
R =
x
A
2
̂
D =
[
B
x
A
x
]
̂
R
[
x B +
̂
D(A)
]
= 0
1
A
2
x
2
B
A
2
x +
[
B
A
1
]
= 0
of 67 88
Boron
Boron is a doping agent for artificial intelligence semiconductor
materials, which are primarily silicon and germanium
But, it stands in relation on the periodic table dierently than the other
principle doping agents, phosphorus, arsenic and gallium. This results
in some interesting equations.!
of 68 88
Periodic Table of the AI Elements
It is quite interesting that we can pull out the core artificial intelligence elements from the
periodic table of the elements in such a way that it allows for a three by three matrix:"
"
Because we can number them according to their properties:"
"
13 is an element in row 1 and must be boron because it has three valence electrons. 14 is carbon because
it is in row 1 and has four valence electrons. And so on. So, instead of element Si we have E_24, and
instead of element Ge, we have E_34, and so on. B and P and Ga and As are doping agents for Si and Ge.
Three by three matrixes are the most important because they are associated with vectors in three
dimensional space, and thus are used for taking the cross product between two vectors.
00001: The properties of artificial intelligence substances form the basis of Universal Structure.
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
of 69 88
Asymmetry in AI Elements
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon (Si) and
germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic (As) have an
asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C) and as such have 4
valence electrons. Thus to have positive type silicon and germanium, they need doping agents from group
13 (three valence electrons) like boron and gallium, and to have negative type silicon and germanium they
need doping agents from group 15 like phosphorus and arsenic. But where gallium and arsenic are in the
same period as germanium, boron is in a different period than silicon (period 2) while phosphorus is not
(period 3). Thus aluminum (Al) is in boron’s place. This results in an interesting equation.
The differential across germanium crossed with silicon plus the differential across silicon crossed with
germanium normalized by the product between silicon and germanium is equal to the boron divided by
the average between the germanium and the silicon. The equation has nearly 100% accuracy:
We found (Beardsley, Mathematical Structure, 2020) that the dierential across silicon (P-Al)
times germanium (Ge) over boron (B) plus the dierential across germanium (As-Ga) times
silicon (Si) over boron (B) was equal to the harmonic mean between Si and Ge. This was
interesting because aluminum is used as what I called a dummy doping agent element, which
when inserted predicts the actually doping agent boron, that seems out of place in the periodic
table where the core artificial intelligence elements are concerned. This is written:"
I find it interesting that the AI elements can be pulled out in a 3X3 matrix because the AI elements seem
to be core to everything in the Universe. I wrote earlier:
00002: It may be we can only understand biological life relative to some other construct, like artificial
intelligence.
Si(A s Ga) + Ge(P Al )
SiG e
=
2B
G e + Si
28.09(74.92 69.72) + 72.61(30.97 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(As G a) +
G e
B
(P Al ) =
2 SiGe
Si + Ge
of 70 88
Stokes Theorem states:
We know the harmonic mean H of a function is
And, that the arithmetic mean A of a function is
We have
S
( ×
u ) d
S =
C
u d
r
×
u =
i
j
k
x
y
z
u
1
u
2
u
3
i
j
k
x
y
z
0
Si
B
(Ga)z
Si
B
(As)y
=
Si
B
(As G a)
i
i
j
k
x
y
z
Ge
B
(Al )z 0
Ge
B
(P)x
=
Ge
B
(P Al )
j
u = (u
1
, u
2
, u
3
)
v = (v
1
, v
2
, v
3
)
u = 0
i +
Si
B
(Ga)z
j +
Si
B
(As)y
k
u = 0
i +
Si
B
(Ga)z
j +
Si
B
(As)y
k
H =
1
1
b a
b
a
f (x)
1
d x
A =
1
b a
b
a
f (x)d x
Si
B
(As G a) +
Ge
B
(P Al ) =
Ge Si
Ge
Si
dx
x
of 71 88
But, we want to use Stokes theorem so we want the integral in the numerator. So, we make the
approximation
And, we have
But, this is only 80% accurate. We find it is very accurate if we say
Which yields
We have by molar mass
Thus,…
H A
1
0
1
0
[
Si
B
(As G a) +
Ge
B
(P Al )
]
d x d y
1
Ge Si
Ge
Si
x d x
f (x) =
4
5
x
1
0
1
0
[
Si
B
(As G a) +
Ge
B
(P Al )
]
d x d y
1
Ge Si
4
5
Ge
Si
x d x
Si
B
(Ga) =
28.09
10.81
(69.72) = 181.1688g/mol
Ge
B
(Al ) =
72.61
10,81
(26.98) = 181.2227g/mol
Si
B
(As) =
28.09
10.81
(74.92) = 194.68111g/mol
Ge
B
=
72.61
10.81
(30.97) = 208.02328g/mol
u = 181z
j + 195y
k
v = 181z
i + 208y
k
of 72 88
We can break up our integral into two integrals u, and v:
Holding x constant and y constant in our u and v integrals we have subtle eclipses in the y-z
planes and x-z planes respectively:
1
0
1
0
Si
B
(As G a)d yd z
1
3
1
(Ge Si)
Ge
Si
x d x
1
0
1
0
Ge
B
(P Al )d x dz
2
3
1
(Ge Si)
Ge
Si
yd y
of 73 88
By Molar Mass
We found that the dierential across silicon (P-Al) times germanium (Ge) over boron (B) plus the
dierential across germanium (As-Ga) times silicon (Si) over boron (B) was equal to the
harmonic mean between Si and Ge. This was interesting because aluminum is used as what I
called a dummy doping agent element, which when inserted predicts the actually doping agent
boron, that seems out of place in the periodic table where the core artificial intelligence
elements are concerned. This is written:"
Si
B
(As G a) +
Ge
B
(P Al ) =
2SiG e
Si + Ge
of 74 88
By Density
We ask if the asymmetry in the AI elements in the periodic table due to boron results in a
dynamic equation by molar mass, then does it as well by density? While molar mass is due to the
composition of elements, density is due to the balance between the strong nuclear force holding
protons together balanced by their electric forces that are mutually repulsive.
The density of boron is 2.340 grams per cubic centimeter, that of phosphorus (white
phosphorus) is 1.88 grams per cubic centimeter and gallium is 5.904 grams per cubic centimeter.
of 75 88
Arsenic is 5.7 grams per cubic centimeter, germanium is 5.323 grams per cubic centimeter and
aluminum is 2.7 grams per cubic centimeter. We have the following scenario:
Again we see boron breaks the symmetry in that period three densities are on the same order and
period 4 densities are on the same order, but that of boron is almost the same as silicon in period
three. We see that semiconductor material Si is the the average between doping agent P and
would be doping agent aluminum that takes the place of boron and, that, the average doping
agent Ga and semiconductor material Ge is approximately the average of doping agent As. Thus
we have:
Ga-As=0.204 (differential across Ge)
Al-P=0.82 (dummy differential across Si)
Si~B
Ge/Si~B
(0.82)
Si
B
= 0.816
(0.204)
Ge
B
= 0.464
0.816
0.464
= 1.7586
GeSi = 3.52
3.52
2
= 1.76 1.7586
of 76 88
And, we have
The first factor takes the form of the harmonic mean between a and b, H:
And the second term takes the form of the geometric mean between a and b, G:
The equation is 94.68% accurate.
Aluminum, while a dummy in the equation used to arrive at the dynamics due to asymmetry by
way of boron actually is widely used in AI because it is a conductor, which makes it an electric
shield, so it can be used to enclose electrical circuitry to protect it from electric fields. Thus we
have the two equations by molar mass and density respectively:
But
Can be written
(Al P )
Ga As
B
GeSi
2
Ge
2
Si + P
2
B
2Ge(Ga As)
Si(Al P) + P(Al P)
GeSi
H =
2a b
a + b
G = a b
2(5.323)(5.904 5.7)
2.33(2.7 1.88) + 1.88(2.7 1.88)
(5.323)(2.33) = 2.2155
2.2155
2.340
= 94.68
B
2Ge(Ga As)
Si(Al P) + P(Al P)
GeSi
1
0
1
0
[
Si
B
(As G a) +
Ge
B
(P Al )
]
d x d y
1
Ge Si
4
5
Ge
Si
x d x
B
2Ge(Ga As)
Si(Al P) + P(Al P)
GeSi
of 77 88
But
Is the geometric mean between Ge and Si. The geometric mean between a and b is given by:
Thus our equation in terms of density can be put in integral form as well:
And we see that this integral is correct:
Which is close to
94.5%
B
2Ge(Ga As)
[
Si(Al P) + P(Al P)
]
GeSi
GeSi
G
¯
f = ex p
(
1
b a
b
a
log f (x)d x
)
1
0
1
0
[
B
2Ge(Ga As)
[
Si(Al P) + P(Al P)
]
]
d x d y = ex p
(
1
Ge Si
Ge
Si
log(x)d x
)
5.323(ln(5.323) 5.323) 2.33(ln(2.33) 2.33) = 3.936
3.936
5.323 2.33
= 1.315
e
1.315
= 3.725
(5.323)(2.33) = 3.52
3.52
3.725
= 0.945
of 78 88
By Atomic Radius
I then considered atomic radii of these elements.
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of 80 88
The atomic radii data varies some from source to source. Here I use data set 4 (previous page)
that is the average of respective values in data sets 1, 2, and 3. We have looked at molar mass,
and density and the dynamic relationships that result in terms of them do to the asymmetry
introduced into the AI elements in the periodic table. It is natural to look at radius next. It has a
lot to do with the structure and properties of the elements just like is true of the their molar
masses and densities.
Here we have the differential across Si, times Si/B plus the differential across Ge times Ge/B is
the golden ratio, phi, times the arithmetic mean between Si and Ge in atomic radius.
(Al-P)=143-116=36
(Ga-As)=19
(36)(115/88)=36(1.3)=46.8
19(123/88)=19(1.4)=26.6
46.8+26.6=73.4
(115+123)/2=119
The golden ratio and the golden ratio conjugate are the solution of the quadratic
that meets the conditions and a=b+c"
Where and , .
We have already said
Thus by radius the integral form of the equation is:
119
73.4
= 1.62 Φ = 1.618
(Al P )
Si
B
+ (G a As)
Ge
B
= Φ
Si + Ge
2
(
a
b
)
2
a
b
1 = 0
a
b
=
b
c
Φ =
a
b
ϕ =
5 1
2
ϕ =
1
Φ
¯
f =
1
b a
b
a
f (x)d x
of 81 88
1
0
1
0
[
(Al P )
Si
B
+ (G a As)
Ge
B
]
d x d y =
Φ
Ge Si
Ge
Si
x d x
of 82 88
The Generalized Equation
Returning to the Asymmetry in the AI Periodic Table:
and by molar mass or
and by density or
and by atomic radius
Are respectively and
And, the ratios
and by molar mass or
and by density or
and by atomic radius
Are, quotients , and , respectively, then if
And, since the geometric mean, arithmetic mean, quadratic mean (root mean square), harmonic
mean are special cases of the generalized mean:
p= 0,1, 2, -1,…
Then the generalized form of our equations is:
Where C is some constant. This is all the following equations:
(As G a)
(P Al )
(Al P )
(Al P )
(Al P )
(Ga As)
ΔE
1
ΔE
2
Si
B
Ge
B
B
2Ge(Ga As)
Si
B
2Ge(Ga As)
P
Si
B
Ge
B
Q
1
Q
2
= (ΔE
1
, ΔE
2
)
Q = (Q
1
, Q
2
)
(
1
n
n
i=1
x
i
p
)
1
p
Q = C
(
1
n
n
i=1
x
i
p
)
1
p
of 83 88
By molar mass.
By density.
By atomic radius.
While this is the arithmetic mean when p=1, and the harmonic mean for p=-1, I have used it as
the geometric mean for p=0, which we can see explodes to infinity at zero. But, since I am a
physicist and not a mathematician, I assume taking the limit as p—->0 is the same as evaluating
it at zero. But, as physicists undergo courses of mathematics, I do understand how to present
equations in terms of the formalities of mathematics. The proper way to treat this as a
mathematician is, the limit as p goes to zero is the geometric mean because
However we can generalize the equation to include the geometric mean without having to take
the limit as p goes to zero using the f-mean. We have:
The power mean is obtained by letting
Thus, our equation becomes:
It is the geometric mean if
Si
B
(As G a) +
Ge
B
(P Al ) =
Ge Si
Ge
Si
dx
x
1
0
1
0
[
B
2Ge(Ga As)
[
Si(Al P) + P(Al P)
]
]
d x d y = ex p
(
1
Ge Si
Ge
Si
log(x)d x
)
1
0
1
0
[
(Al P )
Si
B
+ (G a As)
Ge
B
]
d x d y =
Φ
Ge Si
Ge
Si
x d x
M
0
(x
1
, …, x
n
) =
n
n
i=1
x
i
M
f
(x
1
, x
n
) = f
1
(
1
n
n
i=1
f (x
i
)
)
f (x) = x
p
Q = Cf
1
(
1
n
n
i=1
f (x
i
)
)
of 84 88
00001: The properties of artificial intelligence substances form the basis of Universal Structure.
00002: It may be we can only understand biological life relative to some other construct, like artificial
intelligence.
00003: It is a purpose of biological life (C, N, O, H) to discover the properties of (P, B, Si) so it can make
computing machines which are ultimately necessary to its survival.
00004 That node manufacturing technology (semiconductors) is possible on this planet by way of
naturally occurring substances on this planet (Si, P, B) then the possibility exists that these elements have
such properties by intention outside of ourselves.
00005 That the semiconducting elements found on this planet have the same properties throughout the
Universe, and are in great abundance, would indicate they were not made expressly for humans.
00006: The problem is we don’t know what exists in the microcosmos, we must look at it with itself, we
must use the photon to look at the atom, and it is on the same order of magnitude, so it has to much of an
effect on it for determinism, hence the uncertainty principle saying we can’t know what is there, only the
probability of what is in a given region for a given time.
00007: I can see two scenarios here: One, since there are billions of galaxies, each with billions of stars
the probability for life to occur around a handful of them was high, and the Sun is one of these stars. Or
two, The algorithm put in place was for the Big Bang to create hydrogen, then that condensed into stars
which made from it carbon throughout the universe, and having four valence electrons, it could form with
the hydrogen in long chains, the hydrocarbons, from which DNA synthesized proteins, and life arose
across the universe as a single uniform idea throughout. Elements like silicon have four valence electrons
as well but, are more suitable for making artificial intelligence when doped with phosphorus and boron!
f (x) = log(x)
of 85 88
Masculine And Feminine!
of 86 88
The female sex hormone estrogen (E) is C18H24O2=272.38 g/mol and the male sex hormone
testosterone (T) is C19H28O2=288.42 g/mol. Both are made from cholesterol (Ch)
C27H46O=386.65 g/mol. If we notice that (Ch+T)/E=2.5 and Ge/Si=2.6 we write
(Ch+T)/E=Ge/Si
Results in
T(1-Si/Ge)+E(1-Ge/Si)=Ch(Si/Ge-1)
I interpreted this as the masculine (T) is in inverse relation to the feminine (E), but that the two
add up to a whole (Ch) in that the masculine has coefficient 1-Si/Ge and the feminine has
coefficient 1-Ge/Si that is they are inverse relation but compliment one another. How would an
AI use this information to determine its sex?…
The male is reduced less in the difference between 1 and Si/Ge, but the the female is reduced
less by having Ge in the numerator. It is really quite egalitarian.
of 87 88
Conclusion: The question remains, what is a definition for the weird derivative in general, not
just for e to the x?!
of 88 88
The Author"